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This is sort of a follow up question to this question on Satake-Tits diagrams.

In the question, user Callum comments that for, $\mathfrak{so}(n, \mathbb{C})$, the real forms (potentially with double counting up to algebra homomorphism) are given by $\mathfrak{so}(p,q, \mathbb{R})$ with $p+q=n$, and then if $n$ is even also $\mathfrak{so}^*(n)$. In particular I'm assuming we are considering real Lie algebra's over $M_{n\times n}$ for entries in either $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$.

My question is twofold:

1. It does not seem obvious why there couldn't be $\mathfrak{so}^*(p,q)$. Do these not form a group, or do they always coincide with another real form? ANSWER: Coming back to this now, it is clear Sylvester's law of inertia does not hold for these preserved structures, so $\mathfrak{so}^*(p,q)$ is always a change of basis away from deserving the name $\mathfrak{so}^*(n)$.

2. How do we prove that this is all of the matrix groups which, after complexification, give the relevant group? (This question is not necessarily about the orthogonal groups in particular but real forms of any and all semi-simple Lie algebra's)

In small enough dimension it makes sense to do something like an exhaustive search. However given my surprise at the existence of the quaternionic orthogonal algebra it seems likely counting signature's alone as my intuition tells me, doesn't guarantee I'll find them all, especially as the algebra's get larger.

My current approach is that if there is a signature to play with, each one will be a real form. Then, look to covering maps (in either direction) and see if playing with the signature there gives anything new. Lastly, special cases of quaternionic real forms, and 'scalar restriction' of a complex group. But this is not as systematic as I would like, and leaves me unconvinced I have 'every real form'.

Craig
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  • I think you mean a Lie group rather than a Lie algebra (as your notation suggests). – Moishe Kohan May 08 '24 at 20:48
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    Well, complexification is a messy subject when discussing groups, so algebra's are a clearer and simpler point of discussion. One also does not need to worry about whether one should count different representations as 'different' real forms because the representations have isomorphic real Lie algebra's. – Craig May 08 '24 at 20:55
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    The word "isogeny" applies to Lie groups, not to Lie algebras. – Moishe Kohan May 08 '24 at 21:05
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    I wrote a thesis about this over $p$-adic fields, which I think have a comparable difficulty level to $\mathbb R$. You basically have some criteria about what a Satake-Tits diagram can NOT be. And for the remaining ones, you construct. But this is not done in one afternoon or two. – Torsten Schoeneberg May 08 '24 at 23:28
  • Your thesis wonderfully is in my library but I was hoping maybe when the field was $\mathbb{R}$ perhaps there was a faster or more well known result. I will give it a thorough read through. – Craig May 08 '24 at 23:51
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    To add to Torsten's comment here is one of their answers explaining those criteria on the potential Satake diagrams and why there isn't an easy way to classify them completely. – Callum May 09 '24 at 10:05
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    I would say your approach for hunting for real forms is missing some things, however. I believe you can find all real forms of a classical complex semisimple Lie algebra by considering its defining representation and imposing a compatible real structure, quaternionic structure or Hermitian structure (as discussed, proving this covers all of them is hard) – Callum May 09 '24 at 18:15
  • Yes, I guess the question becomes `how many compatible structures are there' which seems not any easier to answer, but is at least possibly an algebraic question. Is it true that $SO(4,\mathbb{C})$ is the only classical Lie group in the infinite families that is semi-simple and not simple? – Craig May 09 '24 at 21:08
  • @Craig I would argue how many compatible real/quaternionic/Hermitian structures are there (up to isomorphism) is a much more straightforward question to answer. And yes $D_2$ is the only non-simple one in the classical $A_n, B_n, C_n, D_n$ families. You will often see $D_n$ only start from $n=4$ to avoid this, as well as the $D_3 = A_3$ exceptional isomorphism. – Callum May 10 '24 at 08:47

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